Practice Problems: Trig Integrals (Solutions)
[Pages:8]Practice Problems: Trig Integrals (Solutions)
Written by Victoria Kala vtkala@math.ucsb.edu November 9, 2014
The following are solutions to the Trig Integrals practice problems posted on November 9.
1. sec xdx
Note: This is an integral you should just memorize so you don't need to repeat this process again.
Solution:
sec x + tan x
sec2 x + sec x tan x
sec xdx = sec x
dx =
dx
sec x + tan x
sec x + tan x
Let w = sec x + tan x, so du = (sec x tan x + sec2 x)dx:
sec2 x + sec x tan x
1
dx = dw = ln |w| + C
sec x + tan x
w
Plug back in w:
sec xdx = ln | sec x + tan x| + C
2. sec3 xdx Solution: Rewrite:
sec3 xdx = sec x ? sec2 xdx
Use integration by parts. Let u = sec x, dv = sec2 xdx. Then du = sec x tan xdx and v = tan x:
sec3 xdx = sec x tan x - sec x tan2 xdx = sec x tan x - sec x(sec2 x - 1)dx
= sec x tan x - sec3 xdx + sec xdx = sec x tan x + ln | sec x + tan x| - sec3 xdx
Notice on the right side we have the same integral as what we started with, so move it over to the left side:
2 sec3 xdx = sec x tan x + ln | sec x + tan x|
Divide by 2 and add C: sec3 xdx = 1 (sec x tan x + ln | sec x + tan x|) + C 2
1
3. cos4 xdx
Solution: Since we have an even power of cos, we need to use the half angle identity:
cos4 xdx = (cos2 x)2dx =
1
2
1
(1 + cos 2x) dx =
(1 + 2 cos 2x + cos2 2x)dx
2
4
Use half angle again:
1
1
13
1
1 + 2 cos 2x + (1 + cos 4x) dx =
+ 2 cos 2x + cos 4x dx
4
2
42
2
13
1
=
x + sin 2x + sin 4x + C
42
8
4. t sin2 tdt
Solution: Use half angle identity:
t sin2 tdt =
1
1
t (1 - cos 2t) dt =
2
2
tdt - t cos 2tdt
The first integral is straightforward, use integration by parts (tabular method) on the second with u = t, dv = cos 2tdt:
t sin2 tdt = 1
1 t2
-
1 t
sin 2t
-
1
cos
2t
+C
22 2
4
5.
sin3
x
x
dx
Solution: Let w = x, so dw =
1 dx 2dw = 1 dx:
2x
x
sin3
x
dx = 2
sin3 wdw = 2
sin w ? sin2 wdw = 2
x
sin w(1 - cos2 w)dw
Let y = cos w, so dy = - sin wdw:
2 sin w(1 - cos2 w)dw = -2 (1 - y2)dy = -2 y - 1 y3 3
Plug back in w: -2
Plug back in x and add C:
y - 1y3 3
= -2
cos w - 1 cos3 w 3
-2
cos w - 1 cos3 w
= -2
cos
x
-
1
cos3
x
+C
3
3
2
6.
0
sin2
t
cos4
tdt
Solution: You can use half angle identity on this problem, but you would need to use it several
times. I don't think you would see a problem like this on your exam, but it is nice to practice
anyway.
sin2 t cos4 tdt =
sin2 t cos2 t cos2 tdt =
(sin t cos t)2
1 (1 + cos 2t)
dt
0
0
0
2
1
1
2
1
sin 2t (1 + cos 2t)dt =
(sin 2t)2(1 + cos 2t)dt
20 2
80
1 =
(sin 2t)2dt + (sin 2t)2 cos 2tdt
80
0
Let's look at these integrals separately. The left integral we need to use half angle identity:
sin2 2tdt = 1
1
1
(1 - cos 2t)dt = t - sin 2t
=
0
20
2
2
02
Now let's look at the right integral. Use the substitution w = sin 2t, then dw = 2 cos 2tdt:
So the final answer is:
(sin 2t)2 cos 2tdt
=
1
0
w2dw = 0
0
20
sin2 t cos4 tdt = 1
=
0
8 2 16
7.
/2 0
(2
-
sin
)2d
Solution: Multiply this all out and use half angle identity:
/2
(2 - sin )2d =
/2
(4 - 4 sin + sin2 )d =
/2
11
4 - 4 sin + - cos 2 d
0
0
0
22
/2 9
1
9
1
/2 9
=
- 4 sin - cos 2 d = + 4 cos - sin 2 = - 4
0
2
2
2
4
0
4
8. cos2 x sin 2xdx Solution:
cos2 x sin 2xdx = cos2 x ? 2 sin x cos xdx = 2 sin x cos3 xdx
3
Let w = cos x, dw = - sin xdx:
2
sin x cos3 xdx = -2
w3dw
=
1 -
w4
+
c
=
-
1
cos4
x
+
C
2
2
9. tan x sec3 xdx Solution:
tan x sec3 xdx = sec2 x ? sec x tan xdx
Let w = sec x, dw = sec x tan xdx: sec2 x ? sec x tan xdx =
w2dw = 1 w3 + C = 1 sec3 x + C
3
3
10. x sec x tan xdx Solution: Use integration by parts with u = x, dv = sec x tan xdx. Then du = dx, v = sec x:
x sec x tan xdx = x sec x - sec xdx = x sec x - ln | sec x + tan x| + C
11. csc xdx
Note: This is similar to the first problem. This is an integral you should just memorize so you don't need to repeat this process again.
Solution:
csc x - cot x
csc2 x - csc x cot x
csc xdx = csc x
dx =
dx
csc x - cot x
csc x - cot x
Let w = csc x - cot x. Then dw = (- csc x cot x + csc2 x)dx:
csc2 x - csc x cot x
1
dx = dw = ln |w| + C = ln | csc x - cot x| + C
csc x - cot x
w
12. cot3 xdx Solution: cot3 xdx = cot x cot2 xdx = cot x(csc2 x - 1)dx = cot x csc2 xdx - cot xdx
4
= csc x ? csc x cot xdx - cot xdx
Let's look at the first integral. Let w = csc x, then dw = - csc x cot xdx:
csc x ? csc x cot xdx = -
wdw
=
-
1 w2
=
1 -
csc2
x
2
2
Now let's look at the second integral. Rewrite it and let y = sin x so dy = cos xdx:
cos x
1
cot xdx =
dx = dy = ln |y| = ln | sin x|
sin x
y
Now combine the two answers and add C: cot3 xdx = - 1 csc2 x + ln | sin x| + C 2
13. sin 8x cos 5xdx
Solution: I don't think you would see a problem like this on your exam, but it is nice to
practice anyway. There is a trig identity listed on page 476 of your text: sin A cos B =
1 2
[sin
(A
-
B
)
-
sin
(A
+
B)].
You
can
also
derive
this
equation
yourself.
1 sin 8x cos 5xdx =
2
11
1
(sin 3x + sin 13x) dx = - cos 3x - cos 13x + C
23
13
14. cos x cos 4xdx
Solution: Similar to the previous problem, I don't think you would see a problem like this on
your
exam.
On
page
476
of
your
text
is
the
identity
cos A cos B
=
1 2
[cos(A
-
B)
+
cos(A
+
B)].
1 cos x cos 4xdx =
2
1 (cos (-3x) + cos 5x)dx =
2
(cos 3x + cos 5x)dx
11
1
=
sin 3x + sin 5x + C
2 3
5
15.
/6 0
1
+
cos
2xdx
Solution: This is using the half identity backwards.
5
/6
/6
1 + cos 2xdx =
0
0
1 2 ? (1 + cos 2x)dx =
/6
2 cos2 xdx = 2
/6
cos xdx
2
0
0
/6
2
= 2 sin x =
0
2
16.
/4 0
1
-
cos
4d
Solution: This is using the half identity backwards.
/4
/4
1 - cos 4d =
0
0
1 2 ? (1 - cos 4)d =
/4
2 sin2 2d = 2
/4
sin 2d
2
0
0
1
/4
2
= 2 - cos 2
=
2
0
2
17.
1-tan2 sec2 x
x
dx
Solution:
1 - tan2 x sec2 x dx =
(cos2 x - sin2 x)dx =
1 cos 2xdx = sin 2x + C
2
18.
dx cos x-1
Solution: Multiply by the conjugate:
dx
1 cos x + 1
=
?
dx =
cos x - 1
cos x - 1 cos x + 1
cos x + 1 cos2 x - 1 dx =
cos x + 1 - sin2 x dx
= (- csc x cot x - csc2 x)dx = csc x + cot x + C
19. x tan2 xdx Solution: Use the identity tan2 x = sec2 x - 1: x tan2 xdx = x(sec2 x - 1)dx = x sec2 xdx - xdx
6
The last integral is no problemo. The first integral we need to use integration by parts. Let u = x, dv = sec2 x. Then du = dx, v = tan x, so:
x sec2 xdx = x tan x - tan xdx
You can rewrite the last integral as
sin cos
x x
dx
and
use
the
substitution
w
=
cos
x.
- ln | cos x|, so:
x sec2 xdx = x tan x + ln | cos x|
tan xdx =
Plug that into the original integral: x tan2 xdx = x tan x + ln | cos x| - 1 x2 + C 2
20. x sin2 (x2)dx Solution: Let w = x2, dw = 2xdx:
x sin2 (x2)dx = 1
sin2 wdw = 1 ? 1
1
1
(1 - cos 2w)dw = w - sin 2w + C
2
22
4
2
1
1
= (w - sin w cos w) + C =
x2 - sin (x2) cos (x2)
+C
4
4
21. Find the area of the region bounded by the given curves: y = sin3 x, y = cos3 x, /4 x 5/4
Solution: To find the area we need to subtract the bottom function from the top function and then integrate over our domain:
5/4
5/4
5/4
A=
(sin3 x - cos3 x)dx =
sin3 xdx -
cos3 xdx
/4
/4
/4
Let's look at the first integral:
5/4
5/4
5/4
sin3 xdx =
sin x ? sin2 xdx =
sin x(1 - cos2 x)dx
/4
/4
/4
Let w = cos x, dw = - sin xdx:
5/4
- 2/2
sin x(1 - cos2 x)dx = -
(1 - w2)dw =
/4
2/2
= w - 1w3 3
2/2
2
=
- 2/2
2- 6
2/2
(1 - w2)dw
- 2/2
7
Now let's look at the second integral:
5/4
5/4
5/4
cos3 xdx =
cos x ? cos2 xdx =
cos x(1 - sin2 x)dx
/4
/4
/4
Let y = sin x, dy = cos xdx:
5/4
- 2/2
cos x(1 - sin2 x)dx =
(1 - y2)dy
/4
2/2
= y - 1y3 3
- 2/2
2
2/2
=-
2+ 6
Now plug this back in:
A=
5/4
sin3 xdx -
5/4
cos3 xdx =
2
2-
2
- - 2+
2 52
=2 2- =
/4
/4
6
6
3
3
22. Find the volume obtained by rotating the region bounded by the given curves about the specified axis: y = sec x, y = cos x, 0 x /3 about y = -1.
Solution: Use the washer method. The outer area is given by A = r2 = (sec x + 1)2, and the inside area is given by A = r2 = (cos x + 1)2:
/3
/3
V=
(outer area - inside area)dx =
((sec x + 1)2 - (cos x + 1)2)dx
0
0
=
/3
(sec2 x+2 sec x-cos2 x-2 cos x)dx =
/3 sec2 x + 2 sec x - 1 - 1 cos 2x - 2 cos x dx
0
0
22
11
/3
= tan x + 2 ln | sec x + tan x| - x - sin 2x - 2 sin x
24
0
1 2
3
= tan + 2 ln sec + tan - - sin - 2 sin = 2 ln(2 + 3) - -
3
3
3 64 3
3
68
8
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